Problem: Simplify the following expression: $y = \dfrac{3x^2+2x- 8}{3x - 4}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(3)}{(-8)} &=& -24 \\ {a} + {b} &=& &=& {2} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-24$ and add them together. Remember, since $-24$ is negative, one of the factors must be negative. The factors that add up to ${2}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-4}$ and ${b}$ is ${6}$ $ \begin{eqnarray} {ab} &=& ({-4})({6}) &=& -24 \\ {a} + {b} &=& {-4} + {6} &=& 2 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({3}x^2 {-4}x) + ({6}x {-8}) $ Factor out the common factors: $ x(3x - 4) + 2(3x - 4)$ Now factor out $(3x - 4)$ $ (3x - 4)(x + 2)$ The original expression can therefore be written: $ \dfrac{(3x - 4)(x + 2)}{3x - 4}$ We are dividing by $3x - 4$ , so $3x - 4 \neq 0$ Therefore, $x \neq \frac{4}{3}$ This leaves us with $x + 2; x \neq \frac{4}{3}$.